Determine the empirical formula of an oxide of iron which has 69.9% iron by mass.

To determine the empirical formula, we need to find the simplest whole number ratio of iron (Fe) to oxygen (O) atoms in the oxide. Let's assume we have 100g of the iron oxide.

This means we have 69.9g of Fe and (100g - 69.9g) = 30.1g of O.

Next, we convert the masses to moles using the molar masses:

• Moles of Fe = (69.9g) / (55.85g/mol) ≈ 1.25 mol
• Moles of O = (30.1g) / (16.00g/mol) ≈ 1.88 mol

Now, we find the simplest whole number ratio by dividing both mole values by the smaller value (1.25 mol):

• Fe: 1.25 mol / 1.25 mol = 1
• O: 1.88 mol / 1.25 mol ≈ 1.5

Since we need whole numbers, we multiply both values by 2 to get:

• Fe: 1 * 2 = 2
• O: 1.5 * 2 = 3

Therefore, the empirical formula of the iron oxide is Fe₂O₃ (Iron(III) oxide).